2

Value of p such that $\mathop {\lim }\limits_{x \to \infty } \left( {{x^p}\left( {\sqrt[3]{{x + 1}} + \sqrt[3]{{x - 1}} - 2\sqrt[3]{x}} \right)} \right)$ is some finite | non-zero number.

My approach is as follow

$\mathop {\lim }\limits_{x \to \infty } \left( {{x^p}\left( {\sqrt[3]{{x + 1}} + \sqrt[3]{{x - 1}} - 2\sqrt[3]{x}} \right)} \right) \Rightarrow \mathop {\lim }\limits_{x \to \infty } \left( {{x^{\frac{{3p}}{3}}}\left( {\sqrt[3]{{x + 1}} + \sqrt[3]{{x - 1}} - 2\sqrt[3]{x}} \right)} \right)$

$\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt[3]{{{x^{3p}}}}\left( {\sqrt[3]{{x + 1}} + \sqrt[3]{{x - 1}} - 2\sqrt[3]{x}} \right)} \right) \Rightarrow \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt[3]{{{x^{3p + 1}} + {x^{3p}}}} + \sqrt[3]{{{x^{3p + 1}} - {x^{3p}}}} - 2\sqrt[3]{{{x^{3p + 1}}}}} \right)$

How do we proceed

Samar Imam Zaidi
  • 8,800

2 Answers2

3

$$L=\mathop {\lim }\limits_{x \to \infty } \left( {{x^p}\left( {\sqrt[3]{{x + 1}} + \sqrt[3]{{x - 1}} - 2\sqrt[3]{x}} \right)} \right)$$ $$L=\mathop {\lim }\limits_{x \to \infty } \left( {{x^{p+1/3}}\left( {\sqrt[3]{{1 + 1/x}} + \sqrt[3]{{1 - 1/x}} - 2} \right)} \right)$$ Use $(1+z)^k=1+kz+\frac{k(k-1)}{2}z^2+O(z^3)$ when $z$ is very small, then let $z=1/x$ $$L=\lim_{z\rightarrow 0}~ z^{-p-1/3}\left(1+\frac{z}{3}-\frac{z^2}{9}+O(z^3)+1-\frac{z}{3}-\frac{z^2}{9}+O(z^3)-2\right)$$ $$L=z^{-p-1/3}\left(-\frac{2z^2}{9}+O(z^3)\right)$$ Let $-p-1/3+2=0$, then $$L=-\frac{2}{9}$$ Hence $p=\frac{5}{3}$ makes the limit finite equal to $-\frac{2}{9}$.

Z Ahmed
  • 43,235
  • +1 , We were "Simultaneously" giving same type of Answer !! Some Comments : [1] the 3rd line is mixing $x$ & $z$ , & the fifth line is missing a square bracket. [2] I get what you trying to state with $O(z^3)$ , but that looks "Strange" with $O(1/x^3)$ which looks contradictory. – Prem Aug 12 '22 at 19:39
  • OK thanks for your comments I have edited my answer. – Z Ahmed Aug 13 '22 at 04:51
  • I was Pointing to $kz+k(k-1)XXX^2/2$ which must be $kz+k(k-1)ZZZ^2/2$ – Prem Aug 13 '22 at 05:01
2

$\lim \limits_{x \to \infty } {x^{P}(\sqrt[3]{x+1} + \sqrt[3]{x-1} -2\sqrt[3]{x})} = \lim \limits_{x \to \infty } {x^{P}(\sqrt[3]{x}\sqrt[3]{1+1/x} + \sqrt[3]{x}\sqrt[3]{1-1/x} -2\sqrt[3]{x})}$

$\lim \limits_{x \to \infty } {x^{P}(\sqrt[3]{x+1} + \sqrt[3]{x-1} -2\sqrt[3]{x})} = \lim \limits_{x \to \infty } {x^{P}(\sqrt[3]{x}(1+(1/3x)-(1/9x^2) + \cdots) + \sqrt[3]{x}(1-(1/3x)-(1/9x^2) + \cdots) -2\sqrt[3]{x})}$

$\lim \limits_{x \to \infty } {x^{P}(\sqrt[3]{x+1} + \sqrt[3]{x-1} -2\sqrt[3]{x})} = \lim \limits_{x \to \infty } {x^{P}(\sqrt[3]{x}(-(1/9x^2) + \cdots) + \sqrt[3]{x}(-(1/9x^2) + \cdots) )}$

$\lim \limits_{x \to \infty } {x^{P}(\sqrt[3]{x+1} + \sqrt[3]{x-1} -2\sqrt[3]{x})} = \lim \limits_{x \to \infty } {x^{P}(\sqrt[3]{x}(-2(1/9x^2) + \cdots)) }$

$\lim \limits_{x \to \infty } {x^{P}(\sqrt[3]{x+1} + \sqrt[3]{x-1} -2\sqrt[3]{x})} = \lim \limits_{x \to \infty } {x^{P+(1/3)-2}(((-2/9) + \cdots)) }$

The Higher Negative Powers will tend to $0$ in the limit.

The Only Power term we have is $x^{P+(1/3)-2}$ which must be $x^0$ hence ${P+(1/3)-2} = {0}$ ${P+(1/3)-2} = {5/3}$

P less than that will give limit $0$
P more than that will give limit $\infty$

When $P=5/3$ : The Limit is $-2/9$

Prem
  • 9,669